The exact condition for locally free sheaves on a ringed space $(X,\mathcal O_X)$ to be coherent is exactly that $\mathcal O_X$ be coherent.
a) The condition is clearly necessary since $\mathcal O_X$ is locally free.
b) It is sufficient because if the structure shaf is coherent, then coherence is a local property and because a direct sum of coherent sheaves is coherent: apply to $\mathcal F \mid U_i \cong (\mathcal O\mid U_i)^{\oplus r} $
And when is $\mathcal O_X$ coherent?
There is, to my knowledge, no very good non-tautological criterion.
However, for locally noetherian schemes, it is the case that $\mathcal O_X$ is coherent, so for these schemes, yes, locally free sheaves are coherent.
The way I think about this (and it should be isomorphic to the way any reputable source does) is the following:
$L$ is locally equal to $\operatorname{Spec} S(\mathcal{F})$, the spectrum of the symmetric tensor algebra over $\mathcal{F}$. This is pretty easy to understand on an affine patch—you are taking formal sums of tensors of module elements $a+b\otimes c + d\otimes e\otimes f \otimes g + u\otimes w \ldots$, where you quotient out by $x\otimes y - y\otimes x$ so that it's commutative.
Since $\mathcal{F}$ is locally free, it's not too hard to see that $S(\mathcal{F})$ is locally isomorphic—over an open affine patch $\operatorname{Spec} A=U\subset Y$ small enough to trivialize $\mathcal{F}$—to $A[x_1,\ldots,x_n]$, where $n$ is the rank of $\mathcal{F}$.
Let me make that explicit. On $U$, $\mathcal{F}$ looks like the module $M=\bigoplus_{i=1}^n Ax_i$. Then $S(M)$ is exactly the free algebra over $A$ generated by $x_1, \ldots , x_n$. So $S(M) = A[x_1,\ldots,x_n]$, and we can see that it's reasonable to call $\operatorname{Spec} S(\mathcal{F})$ a "vector bundle", since locally it looks like $\mathbb{A}^n_U$.
Okay, now let's suppose that $X\to Y$ is locally given by a morphism of rings $A\to B$, where $\operatorname{Spec} B = V \subset X$. The pullback of $M$ is clearly the base change $M\otimes_A B = \bigoplus_{i=1}^n B x_i$.
But how about pulling back the bundle? Well, it's just $\mathbb{A}^n_U \times_U V = \mathbb{A}^n_V$. Or, on the level of rings, we have $S(M) \otimes_A B = S(M\otimes_A B)$. And that works at the sheaf level, too: $S(\mathcal{F}) \otimes_{\varphi^{-1} \mathcal{O}_Y} \mathcal{O}_X = S(\mathcal{F}\otimes_{\varphi^{-1} \mathcal{O}_Y} \mathcal{O}_X)$.
So this all adds up to a pretty strong dictionary (actually, an equivalence of categories) between locally free sheaves and so-called "geometric vector bundles". Actually, it's a little more general than that—if you look closely at how morphisms of sheaves lift up to the corresponding bundles, you find that as long as $\mathcal{F}$ is reflexive—that is, the natural map $\mathcal{F}\to\mathcal{F}^{\vee\vee}$ is an isomorphism—it can pretty much be identified with its bundle—the sheafy $\operatorname{Spec} S(\mathcal{F})$. You can think of these as singular bundles.
Best Answer
Let $U:=Spec(A)$ and $F:=A\{e_1,..,e_1\}$ the free $A$-module on $e_i$ with $F^*:=A\{x_1,..,x_n\}$. Let $\mathcal{F}$ be the sheafification of $F$.
Question: "But what is $Sym(F∗)(U)$?"
Answer: $Sym_A^*(F^*)(U)\cong A[x_1,..,x_n]$ is the polynomial ring on $x_i$ over $A$.
There is a canonical map
V2. $\pi: \mathbb{V}(F^*):=Spec(Sym_A^*(F^*)):=\mathbb{A}^n_X \rightarrow X$
defined by the canonical map
V3. $\phi: A \rightarrow A[x_1,..,x_n]:=B$
and by definition $\pi^*\mathcal{F}$ is the sheafification of the $B$-module $F\otimes_A B:=F\otimes_A A[x_1,..,x_n]$.